$\dfrac{ 10e + 3f }{ -10 } = \dfrac{ -7e + 4g }{ -10 }$ Solve for $e$.
Answer: Notice that the left- and right- denominators are the same $\dfrac{ 10e + 3f }{ -{10} } = \dfrac{ -7e + 4g }{ -{10} }$ So we can multiply both sides by $-10$ $-{10} \cdot \dfrac{ 10e + 3f }{ -{10} } = -{10} \cdot \dfrac{ -7e + 4g }{ -{10} }$ $10e + 3f = -7e + 4g $ Combine $e$ terms on the left. ${10e} + 3f = -{7e} + 4g$ ${17e} + 3f = 4g$ Move the $f$ term to the right. $17e + {3f} = 4g$ $17e = 4g - {3f}$ Isolate $e$ by dividing both sides by its coefficient. ${17}e = 4g - 3f$ $e = \dfrac{ 4g - 3f }{ {17} }$